3
$\begingroup$

Let $X$ be a topological space. For every point $x \in X$ let $R_x$ be a local ring. Under what (necessary / sufficient / necessary and sufficient) conditions is there a sheaf ${\cal O}_X$ such that $(X,\mathcal{O}_X)$ is a locally ringed space with $\mathcal{O}_{X,x} \cong R_x$ for all $x \in X$?

If $X$ is discrete, we need no conditions, and we can take the sheaf $\mathcal{O}_X(U)=\prod_{x \in U} R_x$. But in general, a necessary condition is that $x \prec y$ gives a ring homomorphism $R_x \to R_y$, and that these are compatible in the sense that $x \mapsto R_x$ extends to a functor from the specialization preorder of $X$ to the category of rings. We will also have $\mathcal{O}_X(U) \subseteq \varprojlim_{x \in U} R_x$, but probably no equality.

$\endgroup$
3
  • 1
    $\begingroup$ Do you know if your necessary condition is sufficient in the following case?: $(X,\leq)$ is a finite partially ordered set and the open sets are the sets $A$ such that if $x\in A$ then $y\in A$ for all $x<y$. I was trying to find counterexamples but I didn't success. $\endgroup$ Apr 26, 2013 at 11:13
  • 3
    $\begingroup$ Why do you ask this question? Is there a specific assignment of local rings that you are trying to "realize"? $\endgroup$ Apr 26, 2013 at 13:46
  • $\begingroup$ The background is a little bit longer, I will send you a mail if you are interested. $\endgroup$ Apr 26, 2013 at 13:56

1 Answer 1

2
$\begingroup$

Here is another necessary condition for the existence of such a sheaf:

For all $x \in X$ and for all $f \in R_x$, there exists a neighborhood $U$ of $x$ such that the $f$ is contained in the image of the canonical homomorphism $$\varprojlim_{y \in U} R_y \to R_x.$$

For suppose that such a sheaf $\mathcal{O}$ exists and let let $f \in R_x$. By definition of the stalk $\mathcal{O}_x$, there exists a neighborhood $U$ of $x$ such that $f$ is in the image of $\mathcal{O}(U) \to \mathcal{O}_x \cong R_x$. But as you indicate above, this factors through the natural map $\varprojlim_{y \in U} R_y \to R_x$, whence $f$ is in the image of the latter homomorphism.

I can't see at the moment whether this is a sufficient condition. My temptation would be to assume that this condition holds and construct a (pre?)sheaf $\mathcal{O}$ by $\mathcal{O}(U) = \varprojlim_{x \in U} R_x$. (Even if this isn't a sheaf, which I'm too lazy/busy to check right now, its sheafification will have the same stalks.) Is this condition enough to ensure that $\mathcal{O}_x \cong R_x$? If not, is there some simple condition that may be added to guarantee this isomorphism?

Edit: As pointed out by Martin in the comments, assuming the condition above, the sheaf defined by $\mathcal{O}(U) = \varprojlim_{x \in U} R_x$ is constructed in such a way that there are natural surjections $\mathcal{O}_x \twoheadrightarrow R_x$. To make the necessary condition sufficient, we only need to ensure that these surjections are also injective. This can be prhased by demanding, for all $x$, that $$\varinjlim_{\mbox{open }U \ni x} \left( \varprojlim_{y \in U} R_y \right) \to R_x$$ is an isomorphism.

Alternatively, injectivity of the maps may be obtained via the following condition (which is also necessary):

For all $x \in X$ and every open neighborhood $U$ of $x$, if $g \in \varprojlim_{y \in U} R_y$ is in the kernel of $\varprojlim_{y \in U} R_y \to R_x$, then there exists a neighborhood $V \subseteq U$ of $x$ such that $g$ is in the kernel of the natural map $\varprojlim_{y \in U} R_y \to \varprojlim_{y \in V} R_y$.

$\endgroup$
3
  • 1
    $\begingroup$ Thank you. It's obviously a sheaf (if $x \prec y$ in $\cup_i U_i$, then $x \in U_i$ for some $i$, but then also $y \in U_i$, etc.) $\endgroup$ Apr 26, 2013 at 19:18
  • 1
    $\begingroup$ Your condition states that $\mathcal{O}_{X,x} \to R_x$ is surjective. Injectivity comes for free in the case of preorders, but I think in general we just have to add it as a condition. So it seems to me that $\varinjlim_{x \in U \text{ open}} \varprojlim_{y \in U} R_y \cong R_x$ is a necessary and sufficient condition. Do you agree? $\endgroup$ Apr 26, 2013 at 19:24
  • $\begingroup$ Martin, yes, that seems correct to me! Though I wonder whether injectivity can be "built in" via some more elementary statement? $\endgroup$ Apr 26, 2013 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.